How to solve this weird-looking equation? So, when I was posting an answer to another user's question, I came up with this strange-looking equation whose roots I want to find. The main problem I have is that the radical sign is really throwing me off, and I'm not sure how I should deal with it. Can someone help me out? (Also, I'm not sure what tag to put this under. I don't think this technically counts as a polynomial, does it?) $$y=\frac {-8\pi x} {\sqrt {2704-4x^2}}+2\pi \sqrt {2704-4x^2}+4\pi x$$ 
 A: Since you only care about roots. Consider
$$\frac {-8\pi x} {\sqrt {2704-4x^2}}+2\pi \sqrt {2704-4x^2}+4\pi x=$$ 
$$\frac{-8\pi x+2\pi(2704-4x^2)+4\pi x\sqrt{2704-4x^2}}{\sqrt{2704-4x^2}}$$
This is 0 if the numerator is 0. So we are solving
$$-8\pi x +2\pi(2704-4x^2)+4\pi x\sqrt{2704-4x^2}=0$$
move $\sqrt{ }$ over and square to get
$$(8\pi x -2\pi(2704-4x^2))^2=16\pi^2x^2(2704-4x^2)$$
This is a degree 4 equation which you can try and solve by factoring if possible or using the awful Cardano formulas. I have left out a fair amount of equivalence checking (denominator 0 for discarding it, and positive/negative values for squaring) but that's just a little more work I think. 
A: Try multiplying both sides with multiplicative inverse of $1/ ( 2704 - 4r^2 ) ^ {1/2}$, it should reduce to a quartic equation in $r$.
A: You could put $y-4\pi x$ on the left-hand side, then square both sides.  Any solution of your equation will also be a solution of the new equation, which has no square-roots any more.
I'm not sure what to do next.
$$(y-4\pi x)^2=\frac{64\pi^2x^2}{2704-x^2}-32\pi^2 x+4\pi^2(2704-x^2)$$
